The book is slightly wrong, but let's try to clear it up. The thing is, you can put 125 on the address bus. But you cannot read addresses 125 and 126 in a single clock by doing so. Basically this is because the numbers 125 and 126 differ in bits other than the last bit.
The 8086 has 20 address lines (A19:A0) and 16 data lines (D15:D0). Actually the 16 data lines and the lower 16 address lines are the same pins, also called AD15:AD0. They function as address lines for the first part of the clock cycle, and data lines for the later part. Also important for addressing is a pin called BHE, "bus high enable". A0 also functions as an "enable" pin, as we'll see in a moment.
If the 8086 wants to read the word at addresses 124-125, It puts 124 on A19:A0, and sets BHE to low. Both banks of memory see the address on A19:A1, and since A0 and BHE are both active-low signals, both banks will be active. The upper bank will put data on D15:D8, and the lower bank will put data on D7:D0, and the CPU gets a full word.
If the 8086 wants to read a single byte at address 124, it puts 124 on A19:A0, and sets BHE high. Because A0 is low (active), the lower bank will put data on D7:D0, but with BHE high, the upper bank does nothing.
If the 8086 wants to read a single byte at address 125, it puts 125 on A19:A0, and sets BHE low. Because A0 is high, the lower bank is inactive, and because BHE is low, the upper bank will put data on D15:D8.
And those are the three cases that are possible (A1 high and BHE high doesn't enable any banks, so it doesn't do anything). Note that the lines A19:A1 are the same in all three cases, and they don't change throughout a clock cycle. If the 8086 wants to read a word at addresses 125-126, it can't get it this way, because those two addresses have different bits on A19:A1. It needs to make two reads, as the book describes.
Your book is considering only A19:A1 as being part of "the address", and A0 as being strictly an enable line, which means it always considers bus addresses to be even. This is basically the truth, since the even and odd bytes of a word are addressed as a unit, but it doesn't really agree with the Intel-approved terminology.